Question

An experiment is run to study if vigorous activities increase blood volumes for healthy men. A group of 7 men participate in vigorous physical activities, and another group of 10 men participate only in everyday, regular activities. Each man has his blood volume (in milliliters) measured after the 1 month experiment. The data are reported below.

Activity Levels

Vigorous : 1612 1352 1456 1222 1560 1456 1924

Regular : 1082 1300 1092 1040 910 1248 1092 1040 1092 1288

(a) [8 marks] Do healthy men with vigorous activities tend to have higher blood volume on average than those with regular activities? Carry out a test at α = 0.05 level of significance. Clearly write down the hypotheses, test statistic, and critical region for the test. What assumptions are needed for the validity of the test?

(b) [8 marks] Build a two-sided 95% confidence interval to assess if it is appropriate to assume that the variances (or variability) of blood volumes are the same for men having vigorous activities and men having regular activities.
For both (a) and (b), clearly give the steps and state the conclusions.

Answer 1

a.
Given that,
mean(x)=1511.7143
standard deviation , s.d1=222.8675
number(n1)=7
y(mean)=1118.4
standard deviation, s.d2 =123.6835
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.943
since our test is right-tailed
reject Ho, if to > 1.943
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =1511.7143-1118.4/sqrt((49669.92256/7)+(15297.60817/10))
to =4.235
| to | =4.235
critical value
the value of |t α| with min (n1-1, n2-1) i.e 6 d.f is 1.943
we got |to| = 4.23495 & | t α | = 1.943
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:right tail - Ha : ( p > 4.235 ) = 0.00274
hence value of p0.05 > 0.00274,here we reject Ho
ANSWERS
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T test for difference of means with unkown population standard deviations
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 4.235
critical value: 1.943
decision: reject Ho
p-value: 0.00274
we have enough evidence to support the claim that healthy men with vigorous activities tend to have higher blood volume on average than those with regular activities.
b.
TRADITIONAL METHOD
given that,
mean(x)=1511.7143
standard deviation , s.d1=222.8675
number(n1)=7
y(mean)=1118.4
standard deviation, s.d2 =123.6835
number(n2)=10
I.
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (6*49669.923 + 9*15297.608) / (17- 2 )
s^2 = 29046.534
II.
standard error = sqrt(S^2(1/n1+1/n2))
=sqrt( 29046.534 * (1/7+1/10) )
=83.989
III.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and value of |t α| with (n1+n2-2) i.e 15 d.f is 2.131
margin of error = 2.131 * 83.989
= 178.981
IV.
CI = (x1-x2) ± margin of error
confidence interval = [ (1511.7143-1118.4) ± 178.981 ]
= [214.334 , 572.295]
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DIRECT METHOD
given that,
mean(x)=1511.7143
standard deviation , s.d1=222.8675
sample size, n1=7
y(mean)=1118.4
standard deviation, s.d2 =123.6835
sample size,n2 =10
CI = x1 - x2 ± t a/2 * sqrt ( s^2 ( 1 / n1 + 1 /n2 ) )
where,
x1,x2 = mean of populations
s^2 = pooled variance
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 1511.7143-1118.4) ± t a/2 * sqrt( 29046.534 * (1/7+1/10) ]
= [ (393.314) ± 178.981 ]
= [214.334 , 572.295]
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interpretations:
1. we are 95% sure that the interval [214.334 , 572.295]contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion